有人可以帮我找到一个更好的替代这个Javascript for循环我写的吗？

Question

Thanks in advance to anyone who can think of a more efficient or a better way to do what my Javascript code below does:

var availableCharacters=Array("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9");

for (counter=0; counter<availableCharacters.length; counter++){
  if(availableCharacters[counter]=="i"||
     availableCharacters[counter]=="l"||
     availableCharacters[counter]=="I"||
     availableCharacters[counter]=="L"||
     availableCharacters[counter]=="1"||
     availableCharacters[counter]=="0"||
     availableCharacters[counter]=="O"){
      availableCharacters.splice(counter, 1);
    }
}

What I'm trying to do is run through an array and remove any elements in that array that are "i", "l", "I", "L", "1", "0" or "O". Whilst this does work it seems like it might be slow and a bit cumbersome. If there is a better way? If not then not a problem, but most of the time when I do something that doesnt seem right to me, it's not! So I thought I'd ask SO

Thanks :)

Answer 1

More recent browsers support Array.filter :

var availableCharacters = ........;
availableCharacters = availableCharacters.filter(function(a) {
  return !a.match(/[ilLI10O]/);
});

For older browsers, however, the for loop given by Mark Linus is good.

Answer 2

var availableCharacters=["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];
for (counter=0; counter<availableCharacters.length; counter++){
    if(/[ilIL10O]/.test(availableCharacters[counter])){
        availableCharacters.splice(counter, 1);
    }
}

Answer 3

Alternative solution with simple filter list:

var availableCharacters=Array("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "0", "1", "2", "3",     "4", "5", "6", "7", "8", "9");
var filter = ['i', 'l', 'I', 'L', '1', '0', 'O'];
for (counter=0; counter<availableCharacters.length; counter++){
    if(filter.indexOf(availableCharacters[counter]) >= 0) {
        availableCharacters.splice(counter, 1);
    }
}

Answer 4

If you're willing to use jQuery you can use grep() and write it like this:

Fiddle here

var availableChars = ["a", "b", "c", '1', 'i', 'o', "9"];
var result = $.grep(availableChars, function(c) {
    return !c.match(/[^il10o]/i)
})
document.write(result)​

Answer 5

filter is nice, but a simple string replace works without a net-

   var availableCharacters=["a", "b", "c", "d", "e", "f", "g", "h", "i", 
   "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", 
   "x", "y", "z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];

  var allowed= availableCharacters.join('').replace(/[iLl10O]/g,'').split('');

  returned value: (Array)
  a,b,c,d,e,f,g,h,j,k,m,n,o,p,q,r,s,t,u,v,w,x,y,z,2,3,4,5,6,7,8,9